Pressure in stars from photons vs. electron/positron pairs

[PAllen]

In a recent popular science account of "pair instability" supernova, a statement was made, without explanation, that if fusion of oxygen started producing sufficiently energetic photons that most of them convert to electron positron pairs, the the outward pressure is drastically reduced and the star collapses. My hope is to get that explanation here. Why do electrons and positrons produce so much less pressure than photons? Further, why wouldn't they annihilate back to photons? (it would seem that in 100+ solar mass star, but not a neutron star, the probability of at least the positrons annihilating would near 1).

Some possible guesses I've made:

In converting much of the photons energy to mass of the electron/positron, the efficiency of momentum transfer to nucleons is much reduced; but I have done no calculations to try to justify this, and it is not obvious to me.

If a positron annihilated, the resulting photons would still have enough energy to convert right pack to a pair.

Any further explanation would be greatly appreciated.

 

[Ich]

Why do electrons and positrons produce so much less pressure than photons?

Pressure has to do with momentum. The more momentum hits a virtual surface in a given time, the more pressure acts on the surface.
Photons carry momentum. If two photons with just enough energy created an electron positron pair, the particles would be at rest, carrying no momentum at all. The energy is in rest mass rather than in momentum then.

why wouldn't they annihilate back to photons?

They do. There woud be an equilibrium, if the conditions lasted that long. The point is: with the onset of pair production, you start draining pressure from the gas, which is exactly what you can't afford if you're desperately fighting to keep the core stable.

 

[twofish-quant]

Why do electrons and positrons produce so much less pressure than photons?

One very crude but hopefully accurate way of explaining it is that you have a gamma ray that is moving at the speed of light and it's bounding around. Now that's convert the gamma rays into electrons and positrons which are moving at much less the speed of light, so there is less bounding around and less pressure.

Further, why wouldn't they annihilate back to photons?

They would. The reaction goes in two directions

e+ + e- <-> 2 gamma rays

When the energy is temperature the reaction goes in one direction because creating electron/positron pairs eats up energy. If you have high temperature then you end up with a lot of "left over energy" which makes the reaction more eve.

In converting much of the photons energy to mass of the electron/positron, the efficiency of momentum transfer to nucleons is much reduced; but I have done no calculations to try to justify this, and it is not obvious to me.

Lets set c=1

E^2 = p^2 + m_0^2

Before the reaction...

E^2 = p_before^2

After the reaction

E^2 = p_after^2 + m_electron^2

You have less momentum in the electrons than you did in the photons.

If a positron annihilated, the resulting photons would still have enough energy to convert right pack to a pair.

That's pretty much it. Also if the gamma ray has just a little energy, it's probably not likely to find another gamma with enough energy to recombine to form an electron-positron pair. If you have really high temperatures, then once the gamma ray forms, it's really likely to find another gamma ray and form an electron-positron pair.

What this means is that when the temperature goes up, you have more pair production. So you have this runaway

high temp -> more reactions -> less pressure -> contraction -> high temp -> ...

 

[PAllen]

Pressure has to do with momentum. The more momentum hits a virtual surface in a given time, the more pressure acts on the surface.
Photons carry momentum. If two photons with just enough energy created an electron positron pair, the particles would be at rest, carrying no momentum at all. The energy is in rest mass rather than in momentum then.
.

What about conservation of momentum?

 

[Ich]

What about conservation of momentum?

Momentum has direction. The net momentum of all perticles is zero anyway, and is conserved.
Pressure (XX-component) is momentum flux in X-direction plus momentum flux in -X-direction (absolute values repectively).

Net momentum 0 can be 0-0 or 100-100. The first case is pressure 0, the second case is pressure 200.

 

[PAllen]

I see the key thing I missed is that you can't balance momentum and energy with 1 photon to e-p pair, or vice versa. It is always two photons to e-p pair and vice versa, required to conserve both E and p. Then, it is clear that e-p pair carries much less momentum per particle than the photon pair.

Thanks.